What is the equation of the line perpendicular to #y=27/12x # that passes through # (2,1) #?

1 Answer
Mar 3, 2018

suppose,the equation of the line required is #y=mx +c#

Now,slope of the given equation #y=(27/12)x# is #27/12=9/4#

If, our required straight line needs to be perpendicular on the given staright line,then we can say, #m.(9/4) =-1#

So, #m=-(4/9)#

So,we found the slope of our line,hence we can put it and write as,

#y=(-4x)/9 +c#

Now,given that this line passes through the point #(2,1)#

So,we can put the value to determine the intercept,

so, #1=(-4*2)/9 +c#

or, #c=17/9#

So,the equation of our line becomes, #y=(-4x)/9 +17/9# or, #9y+4x=17 graph{9y+4x=17 [-10, 10, -5, 5]} #